Let $(X, d)$ be a metric space, and $\gamma: [a,b]\to X$ be a curve. For any partition $P=\{a=y_0<y_1<\cdots<y_n=b\}$, one can associate to it the lengh of the "inscribed polygon" $$\Sigma(P)=\sum_id(\gamma(y_i), \gamma(y_{i+1}))$$
Then we define the length of the curve to be
$$L(\gamma)=\sup_{P\in \mathcal P}\Sigma(P)$$
where $\mathcal P$ is the collection of all partitions of $[a,b]$.
If the supremum is finite then we call the curve is rectifiable.

We denote: $\|P\|=\max_i|y_i-y_{i+1}|$

Now my question is the proof of the following statement:
$$\lim_{\|P\|\to 0}\Sigma(P)=L(\gamma)$$

The hard part for me to prove the statement is if $P$ and $Q$ are two partitions, with $\|P\|\le \|Q\|$, we only know $$\Sigma(P\cup Q)\ge\max(\Sigma(P), \Sigma(Q))$$ and this won't give me any contradiction when we assume there is a sequence of partitions say $P_i$ with $\|P_i\|\to 0$ and $\Sigma(P_i)\le L(\gamma)-\varepsilon_0$ for some fixed $\varepsilon_0>0$. Anybody can help?

(btw. I thought that this may be similar with the proof of the Riemann sum for integrable function, but there one has the Osilation)